Reconciling an apparent contradiction about Parallel Circuits

[Tom.G]

If the contactor had no resistance at all, wouldn't there be 230 volts at the contactor? In other words, if the resistance of the contactor is .0036-Ohms or less, then couldn't the voltage at the contacts be 229.9 volts? I think I might not understand a key feature of electricity here.

It sounds like you have got it either correct or very close!
Here are some details.

With the contacts closed:

1. If you put one meter lead on L2 and the other lead on L1, you will read the power line voltage, in this case 230V.
2. If you move the meter lead from L1 to the contactor terminal with the resistors you will read about 229.9V, the difference between the 230V and the 0.1 V voltage drop across the contacts.
3. If you move the meter lead from L2 to L1 you will read the voltage across the contacts, about 0.1V.
4. If the contacts Open, you will read almost 230V, the voltage across the contacts.

Comments:
Step 2 above is the voltage across the resistors.
Step 4 will actually be 229.4V because the 3000 Ohm meter allows 0.077A to flow. This 0.077A causes a 0.64V drop across the resistors.

Edit: added image

Cheers,
Tom

 

[CWatters]

If the contactor had no resistance at all, wouldn't there be 230 volts at the contactor? In other words, if the resistance of the contactor is .0036-Ohms or less, then couldn't the voltage at the contacts be 229.9 volts? I think I might not understand a key feature of electricity here.

You are confusing these two things..

a) the voltage on the right hand side of the contactor (about 229.9V) and
b) the voltage drop across the contactor (about 0.1V)

 

[Asymptotic]

I'm still confused by how your equation solving for resistance across the contacts. You say that the resistance across the contacts is 0.1 Volts/27.6 amps. You're saying that the voltage going through the contacts is 0.1 volts because the meter resolution of a Fluke 116 multimeter is 0.1 volts. But what does meter resolution mean? Why is the meter resolution the value we should plug in for the voltage when calculating the resistance going through the contactor?

Not quite. What I'm asking is, how much contact resistance is necessary at 27.6 amps (value of the current flowing through them) in order to develop a 0.1V voltage drop? Meter resolution is 0.1V, so this is the resistance necessary to make the meter display change from 0.0V to 0.1V.

One way to think about meter range and resolution is to compare it to a more familiar measuring device. For example, consider a yardstick.

This "meter's" range is 1 yard. It can't measure anything longer than a yard any more than a 600V meter can measure voltages higher than 600V.

Now to resolution. A yardstick with marks at only the origin point (0) and 1 yard mark has exceptionally poor resolution (the thing being measured either fits between the 0 and 1 yard marks, or it doesn't). Dividing it with marks at 1 foot interval increases resolution, but it still isn't very useful, so increasingly finer division marks are added. How much resolution is necessary depends on the application. A carpenter's yardstick may need a resolution no finer than 1/8th of an inch, while a machinist requires finer resolution, and his may go down to 1/64th inch.

In your case, if voltage drop measured by the Fluke 116 meter reads zero, it doesn't necessarily mean the amount of voltage drop is exactly zero, just that it may lie somewhere between zero, and the smallest division of your scale (0.1V).

 

[fourthindiana]

I have reviewed this thread a few times in the past few days.

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor. Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

What do you people think?

 

[fourthindiana]

Not quite. What I'm asking is, how much contact resistance is necessary at 27.6 amps (value of the current flowing through them) in order to develop a 0.1V voltage drop? Meter resolution is 0.1V, so this is the resistance necessary to make the meter display change from 0.0V to 0.1V.

One way to think about meter range and resolution is to compare it to a more familiar measuring device. For example, consider a yardstick.

This "meter's" range is 1 yard. It can't measure anything longer than a yard any more than a 600V meter can measure voltages higher than 600V.

Now to resolution. A yardstick with marks at only the origin point (0) and 1 yard mark has exceptionally poor resolution (the thing being measured either fits between the 0 and 1 yard marks, or it doesn't). Dividing it with marks at 1 foot interval increases resolution, but it still isn't very useful, so increasingly finer division marks are added. How much resolution is necessary depends on the application. A carpenter's yardstick may need a resolution no finer than 1/8th of an inch, while a machinist requires finer resolution, and his may go down to 1/64th inch.

In your case, if voltage drop measured by the Fluke 116 meter reads zero, it doesn't necessarily mean the amount of voltage drop is exactly zero, just that it may lie somewhere between zero, and the smallest division of your scale (0.1V).

I appreciate your efforts to help me understand this. However, let me be frank with you.

The only significance of the concepts of voltage drop and meter resolution that you've said so far is that the two concepts are a useful troubleshooting tool. You said that know voltage drop helps one determine what condition that the contacts are in. I've read your posts on this thread several times both to try to understand them and to try to understand how voltage drop and meter resolution relate to the OP of this thread. To me, voltage drop and meter resolution don't seem germane to the topic of this thread. All this stuff you're writing about voltage drop and meter resolution seems like much ado about nothing, at least to me.

 

[jim hardy]

What do you people think?

I think you're getting mighty close...

In the spirit of 'excruciating attention to detail in wording..'

I have reviewed this thread a few times in the past few days.

Great ! We take that as a compliment .

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

I'm not familiar with the term "Series Laws". Do you mean Kirchoff's Current Law ?

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

Yes ! That's the beginnings of circuit analysis.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor.

Well, not quite. It's in series with the parallel combination of the 10 and 50 ohm resistors.

I guess instructor has not yet covered parallel versus series connections ?

Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Amperage through the contactor is the SUM of the currents through the 10 and 50 ohm resistors. That's 27.6 amps as @Tom.G 's image in post 31.

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps 27.6 multiplied by .05-Ohms
Voltage at contactor= 1.15 1.38 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

Yes it could.Lastly
You have to be careful when saying "voltage AT (anyplace)."
That's because voltage is always taken as a DIFFERENCE between two places.
That's why voltmeters have two wires.

We get sloppy in terminology and it always confuses newbies.
As i said , Voltage is always a difference
L1 is shown as 115 volts. But 115 volts DIFFERENT from what ?
The answer is "115 volts different from circuit common, which in house wiring is almost always connected to "ground:" .
L2 is also 115 volts different from "ground" but it's always opposite direction from L!, so the difference between L1 and L2 is 230 volts.

If you place voltmeter's black lead on circuit ground and its red lead on L1 you'll measure 115 volts.
If you place voltmeter's black lead on circuit ground and its red lead on L2 you'll measure 115 volts.
If you place voltmeter's black lead on L2 and its red lead on L1 you'll measure 230 volts.
Try it - you need to make that intuitive , the thought process needs to be immediate and automatic.

Likewise - if you're going to say "Voltage AT the contactor.."
you must also say 'to where that difference is measured..' Red lead to contactor, Black lead where - to L1. to L2, to ground, or somewhere else ?
If you say instead "Voltage ACROSS the contactor " you've implied red lead to one side of contactor and black lead to other side of contactor.
Try it on your training simulator - that thinking needs to become an automatic habit.
That preciseness in communication will prevent others from misunderstanding what you mean.

When troubleshooting i always ask for a voltage measurement like this:
"What do you read from L2 to R10's side of the contactor?"
and my guys all know that i mean "Place voltmeter's black lead on "From"(L2) and red lead on "To"( resistor side of the contactor) and tell me what you read."
Always describe your voltage so it's unambiguous exactly what difference you are measuring.

It's a bad habit we pick up from reading tire pressures.
We forget that we're actually measuring the difference between pressures inside and outside the tire.

 

[CWatters]

I have reviewed this thread a few times in the past few days.

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor. Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

What do you people think?

It took me awhile to find a data sheet for a contactor that specified the contact resistance. Hope this link works but it says <0.4 milli ohms (<0.0004 Ohms).

https://www.google.com/url?sa=t&sou...WMAR6BAgAEAE&usg=AOvVaw0N9YpKwqUzGdWJMvhNb-QG

Don't get too worried about what your tutor said. When he says "the voltage at the contactor is 0V" he means "the voltage drop across the contactor is approximately 0V because the resistance is low". The voltage drop you calculated is less than 1% of the supply voltage so its a reasonable approximation. What is the likely tolerance of the resistors in your circuit? That will probably have a far greater effect on the current than the contact resistance.Edit: Sorry I can't get that link to work. When Google finds a PDF for you how do you copy the link? I used "copy link address" and then paste it in but it doesn't work.

 

[jim hardy]

Edit: Sorry I can't get that link to work. When Google finds a PDF for you how do you copy the link? I used "copy link address" and then paste it in but it doesn't work.

here's where it sent me

https://www.tdk-electronics.tdk.com/download/2060308/ac1d068dad91b8308dd031bfb3d2f33e/high-voltage-contactor-pb.pdf

 

[Asymptotic]

I appreciate your efforts to help me understand this. However, let me be frank with you.

The only significance of the concepts of voltage drop and meter resolution that you've said so far is that the two concepts are a useful troubleshooting tool. You said that know voltage drop helps one determine what condition that the contacts are in. I've read your posts on this thread several times both to try to understand them and to try to understand how voltage drop and meter resolution relate to the OP of this thread. To me, voltage drop and meter resolution don't seem germane to the topic of this thread. All this stuff you're writing about voltage drop and meter resolution seems like much ado about nothing, at least to me.

The point about meter resolution is, if voltage across the contacts is less than the smallest value the meter can display, then the meter will read zero volts even though a small voltage may actually be present. To further complicate the issue, a Fluke 116 meter reading of 0.1V could represent an actual value between 0.050V to 0.149V - to know it more closely, a higher resolution meter must be used.

Therefore, I still don't know the answer to the following question: Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

This is how resolution ties into the OP. If contact resistance is lower than a certain value for a given current (the amount of resistance necessary to drop 0.1V), then the meter will read zero voltage across the contacts.

 

[DaveE]

First off, the drawing is absurd. You have 115 volts on each side so the voltage across the elements is zero, yet you show it as 230 volts. Second, you have a variable capacitor shown apparently intended as a closed switch.

1) You may want to consider phase? When you see contactors and 10ohm resistor with numbers like 115, 230, 380, 480, then you should consider the possibility that it is an AC circuit.

2) Variable capacitors should be drawn with an arrow through them (according to me) not a line; like a variable resistor. You are confusing the way schematics are drawn by an electrician with the way they are drawn by pcb designers. For example: https://www.nrc.gov/docs/ML1025/ML102530301.pdf

Really, it's ok the way he described it.

This is a staggeringly simple circuit.

Maybe it is to you, by why would he ask about it if he thought that? Are you trying to find faults with his question, show everyone that you know a lot about about electronics, or are you trying to help educate someone who is learning?
What really is the point of your post? Is your reply going to encourage people like the OP to ask similar questions in the future?

 

[fourthindiana]

The point about meter resolution is, if voltage across the contacts is less than the smallest value the meter can display, then the meter will read zero volts even though a small voltage may actually be present. To further complicate the issue, a Fluke 116 meter reading of 0.1V could represent an actual value between 0.050V to 0.149V - to know it more closely, a higher resolution meter must be used.This is how resolution ties into the OP. If contact resistance is lower than a certain value for a given current (the amount of resistance necessary to drop 0.1V), then the meter will read zero voltage across the contacts.

Good points. Now I see how voltage drop and meter resolution are relevant to the OP of this thread. My apologies for saying that the topics are not germane when meter resolution and voltage drop are germane.

 

[Averagesupernova]

I don't think that I need to mention this, but imagine an analog meter with a full scale reading of 10,000 volts. Does meter resolution mean something to you now when trying to measure a tenth of a volt on the 10 KV scale? Exaggeration often times helps me wrap my head around something.

 

[fourthindiana]

Excellent post, jim hardy.

I'm not familiar with the term "Series Laws". Do you mean Kirchoff's Current Law ?

I don't think that what I call the "series laws" are the same as Kirchoff's Current Law. To me, the "Series Laws" are the following three laws:
1# The total current in a series circuit= the current in the first resistor= the current in the second resistor= the current in the third resistor...
2# The total voltage in a circuit= the voltage of the first resistor + the voltage of the second resistor+ the voltage of the third resistor...
3# the law of resistance for series (which I think is similar to the series law regarding voltage but I don't know for sure)

Well, not quite. It's in series with the parallel combination of the 10 and 50 ohm resistors.

I guess instructor has not yet covered parallel versus series connections ?

My instructor has covered parallel versus series connections, but I just momentarily forgot about this stuff because it's been months since I last used it.

Amperage through the contactor is the SUM of the currents through the 10 and 50 ohm resistors. That's 27.6 amps as @Tom.G 's image in post 31.

Okay. Then the voltage through the contactor is 27.6 amps multiplied by .05-Ohms. Voltage through contactor= 1.38 volts

Lastly
You have to be careful when saying "voltage AT (anyplace)."
That's because voltage is always taken as a DIFFERENCE between two places.
That's why voltmeters have two wires.

I was planning on creating a new thread asking about this. I thought that maybe it was the case that voltage is the difference between two places. Now I don't have to create a new thread regarding this.

 

[jim hardy]

I was planning on creating a new thread asking about this. I thought that maybe it was the case that voltage is the difference between two places. Now I don't have to create a new thread regarding this.

There are LOTS of folks whose thinking on that subject is , for lack of a better word, "muddy" .

It has come up in a lot of threads before.
As @Averagesupernova said

Exaggeration often times helps me wrap my head around something.

here's the exaggeration i use to plant the concept. It's repeated in several old threads

Langauge gets so confusing... because we all take shotcutsI was taught in high school that voltage is potential difference
but what was potential ?
The concept of work done moving against a field just didnt seem to relate .

What clarified it for me was when concept of absolute potential clicked in place -

' the work done in bringing a unit of charge from infinity to wherever you are...'

i could just imagine myself traveling from Alpha Centauri to Miami Central High School with a bucketful of charge and tabulating the force encountered every centimeter along the way ... and adding up all those Force X Distance numbers to arrive at "Work Done" .

Nobody is going to go out to infinity and measure the force on a charge,
or hold a voltmeter lead there for you to make that measurement either,

so

we have to settle for measuring the difference in absolute potentials between two points closer to us, and we call that voltage,
We do not know the absolute potential of either point and cannot measure it
but their difference is readily measured by a voltmeter with two leads of finite length.

'Difference in absolute potentials' shortens to potential difference , aka 'voltage'.

So
In my world, voltage is always a potential difference between two points.
We get sloppy in our communication and do not always define those two points,
one of them is usually a 'circuit common' of some sort which is often called 'ground'
though it may be or may not be connected to Earth .

If that potential difference exists because of current flowing through a resistance, that's a voltage drop.
If that potential difference exists because of, let's say the chemistry going on inside a battery, that's not a voltage drop it's just a voltage (or maybe a voltage rise).

Sorry to be so verbose
but the price of clear communication is excruciating attention to language.

Hope mine was clear .. exaggeration is often a useful tool.

old jim

if you nail that concept now, early in your studies, you'll avoid having to "unlearn" a lot of wrong stuff later on.

Good Luck in your class !

old jim

 

[NTL2009]

Jim Hardy's post#36 covers it all very well. I almost wasn't going to join in, but I think there may be a few things I can add.

I congratulate you on your patience and persistence. It's often hard for some us to remember struggling with these concepts, they become 2nd nature after a while, and sometimes hard to explain to beginners. Adding to that, some of those helping in this thread are approaching it at a high level and in general terms (and confusing you a bit with that), while your instructor is looking at things in a bit too low and simple level, and making a bit too many assumptions/generalizations without explanations. You are getting caught in the middle. Hang in there, you are definitely making progress!

I think you have this straight now, but in the earlier posts, you seemed to be confused by the phrase "Current takes the path of least resistance". That does not mean the single path of the lowest resistance (though it seems to read that way). It means whatever path/paths it can take. And when you have parallel resistors, the path of least resistance is to take all the paths combined. I know some here don't like analogies at this level, but I'm not one of them. Consider two parallel lanes of traffic, one with a 50 mph speed limit, one with a 30 mph speed limit. You would never say that the way to get the most cars down those two lanes is to make them all take the 50 mph lane. Obviously, you get more cars by using both lanes to their limits. Currents divide this way too, taking all the parallel paths it can find.

From your post #34:

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

It's useful to try to put some real world experience to these assumptions. A first pass guess of 0.05 Ohms for the contactor was reasonable - after all, 0.05 is a small number compared to the 10 and 50 Ohm values in that circuit. And you did the math right, assuming 23 amps ( I guess this was just with the 10 Ohm in the circuit?), then 23 Amps x 0.05 Ohms = 1.15 V. Now think about power, Volts * Amps. That would be ~ 26 watts at that contactor. Think of a 25 watt soldering iron, or a 25 watt filament bulb. I'd think that contactor would burn up if it was dissipating that much heat. So later on we learn that the contactor is spec'd at typically less than 0.4 milli-Ohms. So that would be 9.2 milli-Volts @ 23 Amps, so ~ 0.21 Watts - much better!

So now you realize that a contactor that has become dirty or pitted, and has had its resistance increase just slightly, can start burning up pretty fast in a circuit with tens of Amps.

So when your instructor says there is zero volts across the closed contacts (and listen to Jim Hardy - voltage is across a component, current is through a component), he is speaking loosely. It's shorthand for "very low, almost zero, not enough to be concerned about". And it is appropriate for dealing with circuits like this, you can't get bogged down in every detail. We know what someone means when they refer to a "12 Volt Battery" in a car, even though it is ~ 12.75 volts at rest, and about 13.8 V when the alternator is running, and maybe ~ 10V when cranking a cold engine. But we have to know when shorthand is OK, and when we need exact values.

This will all be 2nd nature to you too very soon. Get the basics down, everything builds on that.